Lecture Note on Linear Algebra 14. Linear Independence, Bases and Coordinates

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1 Lecture Note on Linear Algebra 14 Linear Independence, Bases and Coordinates Wei-Shi Zheng, 211 November 3, What Do You Learn from This Note Do you still remember the unit vectors we have introduced in Chapter 1: 1 e 1 =, e 2 = 1, e 3 = (1) 1 They are a set of vectors which are the basis (specially the standard basis) of R 3 In this lecture note, we give a more general definition of basis Basic Concept:Basis( 基 ), standard basis( 标准基 ), coordinate system( 坐标系 ), coordinate vector( 坐标向量 ),coordinate mapping( 坐标映射 ) 2 Basis( 基 ) Definition 1 (linear independence) Let V be a vector space Vectors v 1,, v n V are said to be linearly independent iff (if and only if) implies v v n = = = = The set { v 1,, v n } of vectors is called an independent set if v 1,, v n are linearly independent 1

2 注 : 这里 linear independence 是一个 generalized 的定义, 是指在一般向量空间中的定义 如果向量空间是欧式空间, 就是我们第一章的线性独立的定义 在本学期课程, 线性空间暂时限制在欧式空间上定义 Remarks: 1 Vectors v 1,, v n are said to be linearly dependent if they are not linearly independent, or equivalently, there exists,, R which are not all zero such that v v n = 2 If { v 1,, v n } is a linearly independent set then any subset of it is linearly independent 3 If { v 1,, v n } is a linearly dependent set then any superset of it, that is any set contains { v 1,, v n }, is linearly dependent 4 As in the case of R n, itself is linearly dependent So if one of v 1,, v n is, then v 1,, v n are linearly dependent Definition 2 (basis( 基 )) Let V be a vector space and B = { b 1,, b n } an indexed set of vectors of V Then B is called a basis of V iff (1) B is linearly independent; (2) V = Span B Also define the empty set to be the basis of the trivial space { }, namely a vector space contains only the zero vector Example: For any n Z +, {e 1,, e n } is a basis of R n, which is called the standard basis ( 标准基 ) of R n Example: For any n Z +, {1,, x n } forms a basis of R n [x] It is obvious that R n [x] = Span{1,, x n } On the other hand, if c + x+ + x n = then c = = = by equality of polynomials 2

3 Theorem 3 Let V be a vector space and S = { v 1,, v n } V If V = Span S then some subset of S forms a basis of V Proof 1 If v 1,, v n are linearly independent, then by assumption, S is a basis of V 2 Otherwise, v 1,, v n is linearly dependent So there exists v i S such that v i is a linear combination of v 1,, v i 1, v i+1,, v n Define S n 1 = { v 1,, v i 1, v i+1,, v n } Then we have V = Span S = Span S n 1 Now S n 1 S is either a basis of V or a linear dependent set For the case that S n 1 is linearly dependent, again we can remove some v j S n 1 from S n 1 to obtain S n 2 S n 1 such that V = Span S n 2 Repeat this process and finally, we must obtain a linearly independent subset S k of S such that V = Span S k Thus, S k is a basis of V Theorem 4 If two matrix A and B are row equivalent Then, the columns of A has exactly the same linear dependence relationships as the columns of B Proof As A and B are row equivalent, that is A can be row reduced to a matrix B That is the matrix equations A x = and B x = have exactly the same set of solutions That is the columns of A have exactly the same linear dependence relationships as the columns of B Theorem 5 The pivot columns of a matrix A form a basis for ColA Proof We prove this theorem by Four steps: (1) We need to use the result of the last theorem Let B be the reduced echelon form of A The set of pivot columns of B is linearly independent, because no vector in the set is linear combination of the vectors that precede it (2) Since A is row equivalent to B, the pivot columns of A are linearly independent as well according to Theorem 4 (3) The nonpivot columns of B must be the linear combination of the pivot columns of B (4) For this same reason, every nonpivot column of A is a linear combination of the pivot columns of A Thus the nonpivot columns of A may be discarded from the spanning set for ColA, by the Spanning Set theorem This leaves the pivot columns of A as a basis for ColA 3

4 3 Coordinate Systems We first give the definition of coordinate and then we prove that the representation is unique Definition 6 (coordinate) Suppose B = { b 1,, b n } is a basis for V and x is in V The coordinates of x relative to the basis B are the weights,, such that x = bn The vector R n is called the coordinate vector of x with respect to basis B (B coordinates of x) and is denoted by [ x] B We now prove that the coordinate with respect to basis B is unique Theorem 7 Let V be vector space and B = { b 1,, b n } be a basis of V Then for each x V, there exists a unique x = bn = ( b 1 b n ) R n such that Proof The existence of ( ) T R n follows V = Span B Now suppose that (c 1 c n) T R n and x = c 1b c nb n Then = x x = ( b1 + + bn ) (c 1 b 1 + +c n b n ) = ( c 1) b 1 + +( c n) b n Since b 1,, b n are linearly independent, so c 1 = = c n = That is c i = c i for all i The uniqueness follows 4

5 4 Coordinate Mapping: A Linear Transformation View Given a basis B = { b 1,, b n } of R n, for any x R n, its coordinates are, c 2,, We actually can rewrite the following relations: x = bn = P B [ x] B (2) We call P B the change-of-coordinates matrix from B to the standard basis in R n Note that P B is invertible (because the corresponding linear transformation is injective and P B is a square matrix) So we also have: P 1 B x = [ x] B (3) From what we have learned from matrix theory, we have the following: Theorem 8 Let B = { b 1,, b n } be a basis of R n Then, the linear transformation T : P 1 B x [ x] B is injective (one-to-one) from R n to R n We call P 1 B the Coordinate Mapping Now we wish to generalize the above theorem for a more general vector space V as follows: Theorem 9 Let B = { b 1,, b n } be a basis of V Then, the transformation T ( x) = [ x] B is linear and injective (one-to-one) from V to R n Proof There are two steps to prove this theorem We first need to prove T is a linear transformation Second we need to prove it is one-to-one Step 1: Let u, v be two vectors in V, then we have: v = bn, u = d 1 b1 + + d n bn 5

6 So for any scalar e, f, we have T (e v + f u) = T ((e + fd 1 ) b (e + fd n ) b n ) e + fd 1 = e + fd n a = e + f d 1 d n (4) = e[ v] B + f[ u] B = et ( v) + ft ( u) Step 2: We need to prove T is one-to-one Suppose for any v, u, v = bn, u = d 1 b1 + + d n bn If T ( v) = T ( u), then T ( v u) = That is d 1 =,, = d n, and therefore v u 注 : 下面的概念只要求了解 Definition 1 (isomorphism( 同构 )) In general a one-to-one linear transformation from a vector space V onto a vector space W is called an isomorphism from V to W 5 Change of Basis Definition 11 (change of coordinates matrix, 坐标变换 ) Let V be a vector space of dimension n with bases B = { b 1,, b n } and C = {,, } Then for i = 1,, n, we have It is convenient to write c i = ( b 1 b n )[ c i ] B ( ) = ( b 1 b n )([ ] B [ ] B ) = ( b 1 b n )[C] B The n n matrix [C] B = ([ ] B to B [ ] B ) is called the matrix from bases C 6

7 Theorem 12 Let V be a vector space of dimension n with bases B = { b 1,, b n } and C = {,, } Then for any v V, Proof We have [ v] B = [C] B [ v] C v = ( )[ v] C = ( b 1 b n )[C] B [ v] C On the other hand, v = ( b 1 b n )[ v] B By the uniqueness of coordinates, we must have [ v] B = [C] B [ v] C Remarks: 1 Matrix [C] B is invertible, since its rank is n It is easy to see that [C] 1 B = [B] C So [ v] C = [C] 1 B [ v] B = [B] C [ v] B 2 We know that E = {e 1,, e n } is the standard basis of R n Now let B = { b 1,, b n } be any basis of R n Then Examples: Textbook P274, P275 [B] E = ([ b 1 ] E [ b n ] E ) = ( b 1 b n ) The Feast in the House of Levi, by Veronese 7

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